Calculations on Engineering Economics
Calculations on engineering economics(no later than 100 minutes after start)1. Two machines are available. Machine R has fixed monthly cost of 18,800 & a per unit cost of 24.00. Machine U has a
fixed monthly cost of 14,600 & a per unit cost of 31.00. What is the monthly volume when the total cost is equal?_
Answer2. Use an interest rate of 8.00% per period to find the present worth now of 28,000 at the end of period 12.
Answer3. What is the simple interest rate (not the compound interest rate) for the situation of future worth over 5 periods 1.45 times the present worth?
Answer4. Find the unknown i (using compound interest model) for a loan represented by the cash flow below where the amount of the loan is shown in period 0 and the loan payments are shown in other periods with the loan completely repaid after the last payment. Note no currency sign is presented. Must only use lookup tables for this problem.Year 0 1 2 3 4 5 6 7 8Cash Flow 8299 -1540 -1540 -1540 -1540 -1540 -1540 -1540 0
Answer
m\ f *0° °^u'va'ent vv°rth at the end of year 2 for a cash flow stream of + 14.000 at the end of year 8, -10,000 at the UK o \ ear , and -9000 at the end of year 10 applying an annual interest rate of 6.0%?6. Eleven (11) end of year deposits of 5000 earning 6.0% are planned. What is the amount in an account at the time of the last deposit?7. Applying an interest rate of 2.0%, what is the value and sign of x in the period shown in the table that results in theYear 0 1 2 3 4 5 6 7Cash Flow 0 680 680 340 340 170 170 x 1
Answer8. For the stream of end of period cash flows shown below, find the future worth at the end of the period of the last nonzero cash flow using an interest rate of 7.0% per period.Period 1 2 r? .. £ & r 4 6 *7Cash Flow -500 -200 100 400 700 1000 1300
Answer1. The cost of Process A is fixed annual cost of 128,000 and unit cost of 33.00 while Process B has a fixed annual cost of
142,000 and unit cost of 40.00. At what annual volume is the total cost equal?___
Answer2. Applying an interest rate of 15.00% per period find future worth at the end of period 12 of 26,000 now.
Answer3. The future value is 1.54 times the present worth over 6 periods. What is the simple interest rate? Note this is not the
compound interest rate. _
Answer4. Find the unknown i (using compound interest model) for a loan represented by the cash flow below where the amount of the loan is shown in period 0 and the loan payments are shown in other periods with the loan completely repaid after the last payment. Note no currency sign is presented. Must only use lookup tables for this problem._I Year 0 1 2 3 4 5 6 7 81 Cash Flow 6500 -1449 -1449 -1449 -1449 -1449 -1449 0 0
Answer>. A cash flow stream is +18,000 at the end of year 6, -14,000 at the end of year 9, and -9000 at the end of year 10. Applying an annual interest rate of 7.0%, calculate and report the equivalent worth of this cash flow at the end of year 3.6. End of year deposits into an account earning 3.0% are planned. What is the amount in the account after the last deposit is made if 14 deposits of 5000 are made?Answer7. What is the value and sign of x in the period shown in the table applying an interest rate of 6.0% that results in the Present Worth ('PW'l of the cash flow he zero C0.0Y?I Year 0 1 2 3 4 5 6 71 Cash Flow 0 560 560 280 280 140 140 \
Answer8. For the stream of end of period cash flows shown below, find the future worth at the end of the period of the last nonPeriod 1 2 3 4 5 6Cash Flow -400 -200 0 200 400 600
Answer
Engineering Economics
Name
Course
Institution
Date
Question 1
1. Two machines are available. Machine has fixed monthly cost of 18,800 & per unit cost of 24,000. Machine U has a fixed monthly cost of 14,600
Total cost= fixed costs+ variable costs
Total variable cost = (Variable cost per unit) (Total number of units)
18,800+24x=14,600+31x7x=4200 and x=600
Monthly volume= 600 units
Total cost =18800+ $24.00(600) = $33, 200
The variable costs vary with the level f activity and considered in decision making unlike the fixed costs (Subramaniam & Watson, 2016).
Question 2
PVIFA 12, 8*7.5361
28,000*7.5361= 211,010.8
Question 3
What is the simple interest (not the compound interest rate) for the situation of future worth over 5 periods 1.45 times the present worth
F=PV(1+r) n=
Simple interest= P*i*n
A = P(1 + rt)
A is future value
P-present value
r- rate
t-time
A = P(1 + rt)1.45= (1+5r)5r=0.45 and r= 9%
Question 4
4. Compound interest model for loan and cash flow
FV=PV (1+i) 1/n
Year
0
8299
1
-1540
2
-1540
3
-1540
4
-1540
5
-1540
7
-1540
8
0
r = (FV / PV) 1/n – 1
For the seven years cash outflows are 1540*7= 10,780
r=10780/8299/ ^1/7-11.30^0.125-1=3.32%
PVA= PMT, FV annuity, 7 years
8299=pmt/ FV 8years8299/1540*FV
FV=5.388961
Question 5
5. What is the equivalent worth at the end of year 2 for a cash flow stream of +14,000 at the end of year 8,-10,000 at the end of year 9, and-9000 at the end of year 10 applying an annual interest rate of 6.0%?
Equivalent
Year
0
1
2
3
4
5
6
7
8
9
10
Cash Flow
14,000
-10,000
-9000
PVIF
1
0.9434
0.8900
0.8396
0.7921
0.7473
0.7050
0.6651
0.62741
0.5919
0.5584
PV
8783.77
-5918.98
-5025.55
NPV
-2160.76
Annuity Factor
6.21
6.80
7.36
Equivalent
1414.5
-870.22
-682.81
Year 8= PV (0.06,(8-2),14,000*0.7050=9569.45
Year 9 PV (0.06,(9-2),=-10,000*0.6651=6650.57
Year 10= PV (0.06,(10-2),=-9,000*0.6274=5646.71
Adding the values is -2427.83
Question 6
6. Eleven (11) end of year deposits of 5000 earning 6.0% is planned, what is the amount in an account at the time of the last deposit?
Year
Opening balance
Transaction
Interest
Closing balance
0
1
5000
300
5300
2
5300
5000
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